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JEE Advanced levelSingle correct optionDifficulty 3 of 3pb215

6.85 g of red lead, $Pb X_{3} O X_{4}$ , is treated with excess dilute nitric acid. The brown residue is $PbO X_{2}$ ( $Pb X_{3} O X_{4} + 4 HNO X_{3} 2 Pb (NO X_{3}) X_{2} + PbO X_{2} + 2 H X_{2} O$ ). With Pb = 207 and O = 16, the mass of $PbO X_{2}$ obtained is

A2.39 g
B7.17 g
C2.07 g
D4.78 g

Answer: A

Molar mass of

Pb X_{3} O X_{4}

= 3(207) + 4(16) = 685 g/mol, so moles = 6.85/685 = 0.01 mol. Each mole yields one mole of

PbO X_{2}

(239 g/mol), so mass = 0.01 x 239 = 2.39 g.

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